JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
There is a circular tube in a vertical plane. Two liquids which do not mix and of densities \(d_1\) and \(d_2\) are filled in the tube. Each liquid subtends \(90^o\) angle at centre. Radius joining their interface makes an angle \(\alpha\) with vertical. Ratio \(\frac{{{d_1}}}{{{d_2}}}\) is

- A \(\frac{{1 + cos\alpha }}{{1 - cos\alpha }}\)
- B \(\;\frac{{1 + tan\alpha }}{{1 - tan\alpha }}\)
- C \(\;\frac{{1 + sin\alpha }}{{1 - cos\alpha }}\)
- D \(\;\frac{{1 + sin\alpha }}{{1 - sin\alpha }}\)
Answer & Solution
Correct Answer
(B) \(\;\frac{{1 + tan\alpha }}{{1 - tan\alpha }}\)
Step-by-step Solution
Detailed explanation
Equating pressure at \(A\) \(\left( {R\cos \alpha + R\sin \alpha } \right){d_2}g = \left( {R\cos \alpha - R\sin \alpha } \right){d_1}g\)…
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